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More on Supercooling:
Upon completion of my experiment, I decided to delve into the topic of supercooling a bit more to find out it’s uses and the
basic principles that make it work.
Latent Heat:
Each time a substance changes physical state, energy is either absorbed or released. This energy is called latent heat. Latent heat of
fusion is the latent heat which is absorbed or released when substances change from liquid to solid and vise-versa. Therefore, the latent
heat of vaporization is the latent heat which is absorbed or released when substances change from liquid to gas and vise-versa. For example:
when ice melts, latent heat is absorbed, hence when water crystallizes into a solid, latent heat is released.
Definition:
The specific latent heat of fusion of a substance can also be defined as the amount of heat required in order to manipulate a unit mass
of solid substance into the liquid state without a change in temperature. For example: Ice at 0°C has a specific latent heat of fusion of
334kJ/kg. This indicates that in order to convert 1kg of ice at 0°C to 1kg of water at 0°C, a 334kJ amount of energy must be used up by
the ice. Therefore, while converting the same amount and temperature of water into the same amount and temperature of ice, 334kJ of energy
will be emitted into the environment. The amount of energy can be measured either in calories or in joules: 1 calorie = 4.184 joules.
To Calculate:
To calculate latent heat, the following formula is used: Q = mcΔT, where Q is the amount of heat absorbed or released by a substance
(in Joules); m is the mass of the substance (in kg); c is the specific heat capacity of a substance (in J/kg°C); and ΔT is the change
in temperature (in °C). The change in temperature (ΔT) refers to the temperature that was before the substance was either heated or
cooled and by how many degrees it changed when the substance was heated or cooled. To calculate this, the following equation is used:
ΔT = T2-T1; where T1 is the temperature before it was varied, and T2 is the temperature after it had been altered. The material’s mass
is easily calculated by either weighing it, or converting its volume to mass.
Specific Heat Capacity:
The specific heat capacity of a substance is the amount of heat energy (in Joules) required to change the temperature of 1g of the
substance by 1°C. The specific heat capacity of water is 4.184 J/g°C. This means that to raise 1g of water by 1°C, 4.184J of energy
is required to be inputted into the water. Likewise, if water drops in temperature by 1°C, 4.184J of energy will be given out. To
calculate the specific heat capacity of a substance, the formula: c = Q/mΔT is used.
Calculations:
The formula for calculating latent heat is Q = mcΔT. Here are the values for mcΔT as I found out from my experiment:
m = 500mL or 0.5dm3 of fresh water equals to 0.5kg since density of water is 1000 kg/m3 or 1.0 kg/dm3
c = 4.184 J/g°C or 4184 kJ/kg°C
ΔT = 8°C
Therefore after calculating for 500mL of water, the result is as follows:
Q = mcΔT
Q = 0.5 * 4184 * 8
Q = 16736J = 16.736kJ
From this calculation, it is observed that 16736J of energy will be absorbed when water’s temperature rises from -8°C to 0°C.
Table of Specific Heat Capacities:
This is a compiled table of a few random substances and their heat capacities for comparison with water.
| Substance |
Phase |
Specific Heat Capacity (J/g°C) |
| Water |
Liquid |
4.184 |
| Sodium Acetate |
Liquid |
2.5 |
| Aluminum |
Solid |
0.9 |
| Mercury |
Liquid |
0.139 |
| Hydrogen |
Gas |
14.3 |
| Oil |
Liquid |
2 |
| Lithium |
Solid |
3.582 |
| Ethanol |
Liquid |
2.46 |
Smaller specific heat capacity means that less energy will be required to change the temperature of one substance by the
same amount as the temperature of another substance. From this table it is evident that the liquid sodium acetate and mercury
both have smaller specific heat capacity than water. To see exactly how much latent heat is released or absorbed by sodium acetate
compared to water, one must use the same value for the change of temperature (ΔT), as the calculation involving water. Also, the same
amount of sodium acetate must be used.
Additional Calculations:
To determine the amount of energy released or absorbed by sodium acetate when it changes in temperature by 8°C, the following calculation is done:
Q = mcΔT
Q = 0.5 * 2500 * 8
Q = 10000J = 10kJ
When sodium acetate’s temperature decreases by 8°C, 10kJ of energy will be released. On the other hand, when sodium acetate’s temperature
increases by 8°C, 10kJ of energy will be absorbed.
Final Calculations:
If we apply the amount of energy from the first calculation (16736J), to sodium acetate, the temperature change will be different.
The formula used is: ΔT = Q/mc
Q = 16736J = 16.736kJ
m = 0.5kg
c = 2500 J/kg°C
Therefore after calculating for the same amount (500mL) of sodium acetate, the result is as follows:
ΔT = 16736 / (0.5 * 2500)
ΔT = 13.39°C
From this calculation, it is evident that by applying 16.736kJ of energy on a 500mL amount of sodium acetate, changes the temperature
of the sodium acetate by 13.39°C. This is more efficient than using the same amount of energy on the same amount of water.
The change in temperature will be different if a smaller amount of water or sodium acetate is used. The following will be a
comparison between 50mL of water and 50mL of sodium acetate:
Water:
ΔT = Q/mc
Q = 16736J = 16.736kJ
m = 0.05kg
c = 4184 J/kg°C
ΔT = 16736 / (0.05 * 4184)
ΔT = 80°C
The change in temperature is greater since the same amount of energy that was applied to the 500mL was applied to the 50mL.
By decreasing the volume 10 times, the change in temperature increases exactly 10 times as well.
Sodium Acetate:
ΔT = Q/mc
Q = 16736J = 16.736kJ
m = 0.05kg
c = 2500 J/kg°C
ΔT = 16736 / (0.05 * 2500)
ΔT = 133.89°C
From these results, it is evident that sodium acetate has a much higher change in temperature than water. This means that with the same
amount of energy, sodium acetate’s temperature increase will be higher than that of water’s, meaning that 50mL of sodium acetate will be
hotter than 50mL of water when 16736J of energy are applied to it.
Uses:
This is the reason for which sodium acetate is used in heat pads. It is very stable when supercooled to room temperature,
and releases a greater amount of heat than its contenders. The crystallization point of sodium acetate is 54°C. Even though many
other substances have a lower specific heat capacity, because of sodium acetate’s stability and freezing point, it is used instead
of the others. Sodium acetate heat pads are almost always taken on hiking and camping trips in the winter, mountain climbing, as
well as being used in hospitals.
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